JEE Advanced Chemistry Question (2026) — Solution

Step-by-step solution

From the given statement, LiBH_4 selectively reduces the ester group (-CO_2Et) to an alcohol (-CH_2OH) and leaves the carboxylic acid (-CO_2H) unchanged. Borane (BH_3), on the other hand, is a well-known reagent for the chemoselective reduction of carboxylic acids to alcohols in the presence of esters. Let us analyze the first set of reactions yielding P and Q: The starting material is a 1,2-disubstituted cyclobutane derivative. At C1, the ester group (-CO_2Et) is on a wedge (up) and the acid group (-CO_2H) is on a dash (down). The adjacent C2 has a methyl group. In product P (using LiBH_4), the wedge -CO_2Et is reduced. Thus, the new -CH_2OH group is on the wedge. In product Q (using BH_3), the dash -CO_2H is reduced. Thus, the new -CH_2OH group is on the dash. Comparing P and Q, the stereocenter at C1 is inverted relative to the unchanged stereocenter at C2. Since their relative stereochemistry (cis vs trans relationship between -CH_2OH and -CH_3) is different, P and Q are diastereomers. Now let us analyze the second set of reactions yielding R and S: The starting material is a 1,3-disubstituted cyclobutane derivative. At C1, -CO_2Et is on a wedge and -CO_2H is on a dash. At C3, there is a methyl group. In product R (using LiBH_4), the ester on the wedge is reduced. The resulting -CH_2OH group is on the wedge (cis to the methyl group if we assume methyl is up). In product S (using BH_3), the acid on the dash is reduced. The resulting -CH_2OH group is on the dash (trans to the methyl group). Since R and S represent the cis and trans isomers of a 1,3-disubstituted cycloalkane, they are stereoisomers that are not mirror images. Therefore, R and S are also diastereomers. Thus, P & Q are diastereomers, and R & S are diastereomers.